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n3/2 = O (n2); if n = 500

n3/2 = O (n2); if n = 500

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Exercise on Big-Oh Analysis. Answer True or False whether a function with growth rate is a member of the set of algorithms that grow with a certain complexity. Draw also the graph.

  1.       n3/2 = O (n2); if n = 500
  2.       4n2+5n+2 = Ω (n2); if n = 150
  3.       2n3+8n2+6n+7 = ω (2n); if n = 500
  4.      (n+1)(n-1)/ 3 = o (logn); if n = 200
  5.       5n + 3 = O (nlogn); if n = 1000
  6.  

int[] arr = {2,4,6,8,10,12,14,16};

for(int i= 0; i<arr.length; i++){

System.out.println(arr[i]);

}

= o (n)

7. 

for(int row= 1; row<=5; row++){

for(int col=1; col<=3; col++){

System.out.print(“*”);

}

System.out.println();

}

= Ω (n2)

8.       2n = ω (n); if n = 20

9.      6n2+2n+5 = O (n!); if n = 50

10.    2n + 10 = ϴ (nlogn); if n = 200

Exercise on Big-Oh Analysis. Answer True or False whether a function with growth rate is a member of the set of algorithms that grow with a certain complexity. Draw also the graph. 1. n2 = 0 (n²); if n=500 2. 4n²+5n+2 = Q (?); if n = 150 3. 2n²+8n²+6n+7 = w (2"); if n = 500 4. (n+1)(n-1)/ 3 = o (logn); if n= 200 5. 5n + 3 = 0 (nlogn); if n = 1000 6. int [] arr = {2,4,6,8,10,12,14,16}; for (int i= 0; i<arr.length; i++){ = o (n) System.out.println(arr[i]); } 7. for (int row= 1; row<=5; row++) { for (int col=1; col<=3; col++) { System.out.print ("*"); = 0 (r²) } System.out.println(); } 8. 2" = w (n); if n= 20 9. 6n²+2n+5 = O (n!); if n= 50 10. 2n+ 10 = O (nlogn); if n = 200
Transcribed Image Text:Exercise on Big-Oh Analysis. Answer True or False whether a function with growth rate is a member of the set of algorithms that grow with a certain complexity. Draw also the graph. 1. n2 = 0 (n²); if n=500 2. 4n²+5n+2 = Q (?); if n = 150 3. 2n²+8n²+6n+7 = w (2"); if n = 500 4. (n+1)(n-1)/ 3 = o (logn); if n= 200 5. 5n + 3 = 0 (nlogn); if n = 1000 6. int [] arr = {2,4,6,8,10,12,14,16}; for (int i= 0; i<arr.length; i++){ = o (n) System.out.println(arr[i]); } 7. for (int row= 1; row<=5; row++) { for (int col=1; col<=3; col++) { System.out.print ("*"); = 0 (r²) } System.out.println(); } 8. 2" = w (n); if n= 20 9. 6n²+2n+5 = O (n!); if n= 50 10. 2n+ 10 = O (nlogn); if n = 200
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